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Reaction at the Interfaces Lecture 1

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  • Reaction at the Interfaces

    Lecture 1

  • Lecture Plan

    FPColloids, Biointerfaces and Drug delivery systems9-15

    Thermodynamics of Soft Matter: Glass, Gels and Liquid Crystals.

    8Solid surfaces: Adsorption and Catalysis7

    Solid surfaces: Structure and related experimental techniques.

    6

    Langmuir-Blodgett films: Structure, Physical properties and Applications

    5Langmuir-Blodgett films: Deposition; Phases4

    Adsorption at Gas-Liquid Interface: surfactants and micelles. Self-Assembly of Amphiphilic molecules.

    3

    Thermodynamics of the surfaces and Adsorption. Adsorption at Gas-Liquid interface

    2LGIntroduction. Surface tension. 1

    LecturerTitleLecture

  • Interfaces• interface – the region where properties

    change from one phase to another• particular interest to nanotechnology, as

    disperse phase continuous phase

    2 3surface effects 1bulk effects

    L L L∝ =

  • Interfaces

    all non-uniformity is hier!

    not necessary monotonic change

  • Types of interfaces• it’s possible to classify interfaces based on the nature of

    bulk phase.• Gases intermix completely, so there are no gas-gas

    interface

    1 2

    1 2

    G-Lgas-liquidfluid interfaces

    L -Lliquid1-liquid2

    gas-solid G-Ssolid interfaces liquid-solid L-S

    solid-solid S -S

    ⎧⎨⎩⎧⎪⎨⎪⎩

  • Stability of an interface

    • interface can possess and extra energy, so

    other termsG Aγ= +• surface tension should be positive

    otherwise the system is totally miscible

    • driving processes in surface thermodynamics:– lowering of surface energy through

    minimizing of surface area– lowering of surface energy through

    adsorption on the surface

  • Key concepts

    • Surface tension• Wetting• Adsorption • Emulsions• Colloids• Membranes

  • Surface tension

    • Surface tension can be defined as a force per unit length acting on an interface

    Fx

    γδ

    =

    • Forces can be understood as a result of broken bonds when moved to a dissimilar phase

    • work of extension:

    sw F y x y Aδ γ δ γδ= = =

  • Contact angle, wetting and spreading

    • at equilibrium:

    cosGS LS GLγ γ γ θ= +

    • before equilibrium is reached:

    cos 'h GS LS GLF γ γ γ θ= − −

    • equilibrium shape is a result of balance between cohesive (inside the liquid) and adhesive (on the interface) forces.

    • strong adhesive forces (for water, polar groups leading to hydrophilicity) cause wetting q

  • Contact angle, wetting and spreading• Wetting is determined by the equilibrium contact angle:

    – θ90o – liquid doesn’t wet the solid,– θ=0o – complete or perfect wetting

    cos 'h GS LS GLF γ γ γ θ= − −

    • Spreading coefficient:

    LS GS LS GLS γ γ γ= − −

    compare:

    if SLS > 0 than the liquid spreads completely, otherwise equilibrium contact angle exists

    partial wetting total wetting

  • Work of adhesion and cohesion• zero thickness interface model: surface of tension• consider a process when two phases are pulled apart and two

    new interfaces are created

    wαβ αβ αω βωγ γ γ= − + +work of adhesion:

    2wαα ααγ=work of cohesion:

    • in a case of droplet on solid surface:

    cosGS LS GLγ γ γ θ= +

    LS LS LG SGw γ γ γ= − + +(1 cos )LS LGw γ θ= +

    Young-Dupre

  • Droplet on rough surface: Wenzel’s law

    • assumption: roughness on a microscopic scale

    • work along the contact line:

    *

    ( ) coscos ( )

    ( )cos cos

    x SL SG LG

    LG SL SG

    SL SG

    LG

    dE dW F dx rdx dxr

    r r

    γ γ γ θ

    γ θ γ γγ γθ θ

    γ

    = = = − +

    = −

    −= =

    ∑on smooth surface

    * * 0

    * * 0

    for hydrophobic surface ( 90 ) for hydrophilic surface ( 90 )

    θ θ θ

    θ θ θ

    > >

    < <

    Wenzel’s law

  • Cassie-Baxter law• droplet on an inhomogeneous surface:

    1 1 1 2 2 2

    1 1 2 2

    ( ) ( ) coscos cos cos

    x SL SG SL SG LGdE dW F dx f dx f dx dxf f

    γ γ γ γ γ θ

    θ θ θ

    = = = − + − +

    = +∑

  • Superhydrophobicity and Superhydrophilicity

    • Wenzel’s law: hydrophobocity or hydrophilicity is enhanced by an increase in surface roughness

    • In addition inclusion of air pores contributes to the superhydrophobicity 01 1 2 AIRcos cos as 180f fθ θ θ= − =

  • Measurements of surface tension• Measuring contact angles: sessile or pendent droplet

  • Measurements of surface tension• Wilhelmy plate:

    2( )F x yγ= ⋅ +

    in the past was mainly measured on roughened mica, etched glass etc. Currently paper plates (i.e. filter paper) is the material of choice

    • du Noüy ring:

    length thickness

  • Measurements of surface tension• Capillary rise

    2212

    c c

    c

    r gh r

    ghr

    γ π ρ π

    γ ρ

    = ∆

    = ∆

    alternatively the difference between two capillaries of different diameter can be measured

  • Measurements of surface tension

    • Drop weight (pendent droplet)

    2mg rπ γ=

    • fast and easy• fraction of a drop will stay on the capillary, so a correction

    factor should be applied

  • Measurements of surface tension

    • Maximum bubble pressure

    • sharp edge necessary to define the radius• bubble shape not exactly spherical due to pressure vs. depth

    2Prγ

    ∆ =

  • The Laplace equation

    2

    pressure forces:( ) cos

    ( ) c

    F P P AF P P r

    α β

    α β

    δ δ φ

    π

    = −

    = −

    2( ) (2 ) / 02

    c c cP P r r r r

    P Pr

    α β

    α β

    π γ πγ

    − − =

    − = Laplace equation for spherical surface

    surface tension:(2 )cos (2 ) /z c c cF r r r r

    γ γ π θ γ π=− =−

  • The Laplace equation• in general case:

    1 1 2

    m

    P Pr r r

    α β γγ ⎛ ⎞− = + =⎜ ⎟′ ′′⎝ ⎠

    P1 = P2

  • Measurements of surface tension• Maximum bubble pressure

    1 1 11 1 2

    4 cos 4 4 cosP P Pd d dγ θ γ γ θ− − −

    ∆ = ∆ = ∆ =

    • Valving effect in capillaries

  • The Kelvin equation• vapour pressure above a droplet

    2lnc L

    m

    p Vp RT r

    γ∞

    ⎛ ⎞=⎜ ⎟

    ⎝ ⎠Kelvin equation

    , mdG SdT Vdp Gµ= − + = mT

    Vpµ⎛ ⎞∂

    =⎜ ⎟∂⎝ ⎠

    2 (1/ )m

    V p V pp p r

    α β α α β β

    α β

    δµ δµ δ δ

    δ δ γδ

    = ⇒ =

    − =( ) / 2 (1/ )

    ,/ / 2 (1/ )

    m

    m

    p V V V r

    if V Vp V V V r

    β β α α

    β α

    β β α β α

    δ γδ

    δ δµ γδ

    − =

    = =2 (1/ )

    2ln

    c Lm

    c L

    m

    V r

    p VRTp RT r

    µ µ γ

    γ

    − =

    ⎛ ⎞=⎜ ⎟

    ⎝ ⎠

  • Consequences of Kelvin equation

    • smaller droplets will have higher vapour pressure and therefore evaporate faster

    • small droplets have higher chemical potential• condensation in a capillary• at the phase transition only the nucleation center with

    infinite radius will grow. All finite nucleation center require finite overcooling/overheating (i.e. a thermodynamic force)

    32

    2*

    4 43

    24 8

    m

    m

    m

    rG rV

    VrG r rr V

    π µ π γ

    γπ µ π γµ

    ∆ = − ∆ +

    ∂∆ = − ∆ + =

    ∂ ∆

  • Surface tension of pure liquid

    • surface tension of pure liquids usually decreases nearly linear with temperature

    ( )( )2/3 7 2/3 12.12 10d M J mol KdT

    γ ρ− − −= − ⋅ Eötvös equation

  • Hydrophilic-hydrophobic interaction

    • water properties are governed by the presence of hydrogen bonds

    • polar substances favor such type of arrangement and are therefore hydrophilic

    • non-polar substances lead to clathrate formation and therefore increase free energy (via decrease in entropy)

  • Probelms

    • Problem 2.5 Use the Kelvin equation to calculate the radius of an open-ended tube within which capillary condensation of nitrogen at 77 K and a relative pressure of 0.75 might be expected. Assume that prior adsorption of nitrogen has formed a layer 0.9 nm thick coating the inside of the tube. For nitrogen at 77 K the surface tension is 8.85 mN m-1 and the molar volume is 34.7 cm3 mol-1.

    • Problem 2.6 A jet aircraft is flying through a region where the air is 10% supersaturated with water vapour (i.e. the relative humidity is 110%). After cooling, the solid smoke particles emitted by the jet engines adsorb water vapour and can then be considered as minute spherical droplets. What is the minimum radius of these droplets if condensation is to occur on them and a “vapour trail” form? Data: γ(H2O) = 75.2 mN m-1, M(H2O) = 0.018 kg mol-1, ρ(H2O) = 1030 kg m-3, T = 275 K.